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Does the atmosphere provide a benefit or detriment in terms of fuel economy for rockets that take off and land back on Earth? We know that the atmosphere plays a large role in helping slow down rockets landing on Earth and helps save valuable fuel. Yet the atmosphere also creates lots of resistance on takeoff to the point that some rockets throttle down during periods of high resistance.

Would it take more or less fuel for a reusable rocket such as the Falcon 9 to takeoff and land back on Earth if it had no atmosphere?

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    $\begingroup$ Just to clarify, are you taking the example vehicle as completely unmodified? Very small alterations could make a big difference to fuel-expenditure without an atmosphere - eg. ditch the fairings before launch - but I think that's outside of what you're asking? $\endgroup$
    – Jack
    Aug 7, 2018 at 13:42
  • $\begingroup$ @Jack For this question I was thinking stock vehicle as there are plenty of metrics available and I didn't want it to be too speculative. I do find it interesting though what benefits would be had if you could use the same engine nozzle the whole time, have no fairings and heat shielding, and design the rocket in a shape that purely optimizes weight. $\endgroup$
    – FreakyDan
    Aug 8, 2018 at 13:19

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Let's keep the atmosphere, at the very least so we can continue breathing ;)

What this boils down to is whether there's more delta-V from drag during launch vs. delta-V provided by aerobraking during re-entry. You could argue about the costs associated with heat-resistant panels, but most of your rocket-dollars are going to be put towards other parts of the system.

In a few cases, like with the ISS, you have to consider long-term atmospheric drag, but most objects are in high enough orbits that they're not appreciably losing velocity to drag - after all, we don't re-boost most of them. Satellites will de-orbit after a long enough period of time, but I choose to see that as a perk, given the alternative would be extensive debris junking up the orbital slots we prefer.

Wikipedia tells us we need about 9.4 km/s to reach LEO with an orbital speed of 7.8 km/s. The same section tells us atmospheric and gravity account for the difference, though it ranges between 1.3 and 1.8 km/s. As Russell Borogove points out, most of that is due to gravity and the atmospheric value tops out around a couple hundred

When something comes back into the atmosphere from LEO, it's going to have at least 7.8 km/s of re-entry dV. The vast majority of that is diminished by aerobraking, and the remainder dissipated by chutes, gliding or boosters, depending on the vehicle.

TL;DR: A couple hundred m/s << 7.8 km/s. Atmospheres are helpful

Edit: Changed atmospheric contribution numbers per Russell Borogove's comment

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  • $\begingroup$ I was struggling figuring out how to relate takeoff and landing resistances. The delta-V figures really helped me understand it. Thank you. $\endgroup$
    – FreakyDan
    Aug 7, 2018 at 17:53
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    $\begingroup$ The majority of the loss on ascent is due to gravity rather than atmospheric drag -- it's probably only a couple hundred m/s for atmosphere in most cases. (I know it's about 50 m/s for the Saturn V, but it's generally higher for smaller launchers.) $\endgroup$
    – Russell Borogove
    Aug 7, 2018 at 18:42
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    $\begingroup$ @RussellBorogove Good to know. I guess it makes sense that they'd be optimally aerodynamic, but that's a lot less drag than my non-aerodynamics-aficionado gut feeling $\endgroup$
    – Punintended
    Aug 7, 2018 at 19:50
  • $\begingroup$ Wow this is really great actually. Old post and I would've missed it if not for a question prompt in the "related questions" box of my own question. Really informative! $\endgroup$
    – Anti Elon Guy
    Mar 9 at 1:13

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